Question 12

CaptainBlack · #1 $Old$ December 26th, 2006, 01:31 AM

Water leaves a tap (faucet) in a steady stream at a speed of 20 cm/s
which is 1 cm wide.

How wide is the stream 30 cm below the tap (assuming that the stream
does not break up due to surface tension induced instability).

RonL

galactus · #2 $Old$ December 29th, 2006, 04:03 PM

Hey Cap'n.

No one bit on this problem, so I will give my farthings worth.

I think I have seen a problem like this before, a while back.

I think the width would be proportional to how far it is from the tap.

If I remember right the formula is $R(y)=ky^{\frac{-1}{4}}$

Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.

We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume ${\pi}(R(y))^{2}h$

Because of gravity and all that, the rate this disk is passing y is:

$F={\pi}(R(y))^{2}h\sqrt{19.6y}$

Where F is the rate the water is leaving the tap.

Now, solve for R(y).

Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.

This does look like an interesting little problem, though.

CaptainBlack · #3 $Old$ December 29th, 2006, 04:09 PM

Quote:

Originally Posted by galactus $View Post$

Hey Cap'n.

No one bit on this problem, so I will give my farthings worth.

I think I have seen a problem like this before, a while back.

I think the width would be proportional to how far it is from the tap.

If I remember right the formula is $R(y)=ky^{\frac{-1}{4}}$

Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.

We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume ${\pi}(R(y))^{2}h$

Because of gravity and all that, the rate this disk is passing y is:

$F={\pi}(R(y))^{2}h\sqrt{19.6y}$

Where F is the rate the water is leaving the tap.

Now, solve for R(y).

Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.

This does look like an interesting little problem, though.

The problem of the instability of the stream is more interesting, It is a
problem related to the (in)stability of broad planetary ring systems to break
up into ringlets.

RonL

CaptainBlack · #4 $Old$ December 31st, 2006, 11:54 PM

Quote:

Originally Posted by CaptainBlack $View Post$

Water leaves a tap (faucet) in a steady stream at a speed of 20 cm/s
which is 1 cm wide.

How wide is the stream 30 cm below the tap (assuming that the stream
does not break up due to surface tension induced instability).

RonL

Question 12: Solution

The mass flow rate is a constant so at any point Av is that same
as at any other point, where A is the crossectional area of the stream
and v is the speed of the stream.

A mass element undergoes an accelaration of g m/s^2 (sign convention
is positive downwards), and so at time t its speed is:

v(t)=g.t+v(0)

and it has fallen a distance:

s= g.t^2/2+v(0)t.

So the time to fall 0.3 m, is a root of (taking g=9.81 m/s^2):

0.3=9.81 t^2/2+0.2t,

the roots of this are t~=-0.2685, and t~=0.22776. The first
of these roots is unphysical so we are left with t~=0.22776 seconds.

Its speed when it has fallen 0.3 m is:

v(0.22776)=9.81 0.22776 + 0.2 ~= 2.434 m/s

So now we have from the constancy of the mass flow rate past any point:

pi*(0.01/2)^2*0.2=pi*(D/2)^2*2.434

which gives:

D~=0.002872 m, or D~=0.29 cm.

RonL

topsquark · #5 $Old$ January 1st, 2007, 12:38 PM

Quote:

Originally Posted by CaptainBlack $View Post$

Question 12: Solution

The mass flow rate is a constant so at any point Av is that same
as at any other point, where A is the crossectional area of the stream
and v is the speed of the stream.

A mass element undergoes an accelaration of g m/s^2 (sign convention
is positive downwards), and so at time t its speed is:

v(t)=g.t+v(0)

and it has fallen a distance:

s= g.t^2/2+v(0)t.

So the time to fall 0.3 m, is a root of (taking g=9.81 m/s^2):

0.3=9.81 t^2/2+0.2t,

the roots of this are t~=-0.2685, and t~=0.22776. The first
of these roots is unphysical so we are left with t~=0.22776 seconds.

Its speed when it has fallen 0.3 m is:

v(0.22776)=9.81 0.22776 + 0.2 ~= 2.434 m/s

So now we have from the constancy of the mass flow rate past any point:

pi*(0.01/2)^2*0.2=pi*(D/2)^2*2.434

which gives:

D~=0.002872 m, or D~=0.29 cm.

RonL

Might I point out that this is an application of Bernoulli's equation (an extremely simplified version of fluid dynamics: the flow for a non-viscous incompressible irrotational fluid) along with the constant mass flow-rate condition. I typically assign something like it in my Freshman Physics course.

-Dan